User michael greinecker - MathOverflow

From universal measurability to measurability

2013-05-13T07:21:46Z

Let $(\Omega,\Sigma)$ be a measurable space and $K$ be a compact metrizable space endowed with its Borel $\sigma$-algebra $\mathcal{B}(K)$. Let $A\subseteq\Omega\times K$ be universally measurable and such that $$C_\omega={x\in K:(\omega,x)\in A}$$ is closed for all $\omega\in\Omega$. Let $\Sigma_u$ the universal completion of $\Sigma$. Is it then the case that $A$ is in $\Sigma_u\otimes \mathcal{B}(K)$?

I face the problem when I want to obtain a certain set-valued function that satisfies a strong measurability condition. It is enough for my purposes to get a closed-valued set-valued function with a graph measurable with respect to the completion on my underlying probability space. I can obtain the desired set-valued function as a projection, but this only gives me universal measurability of the graph.

As a first step, it might be intersting to know whether the conjecture holds in the case that $A(\omega)$ contains a single element for all $\omega$, so that it really is a function.

If $f:\Omega\to K$ has a universally measurable graph, is $f$ then $\Sigma_u$-$\mathcal{B}(K)$-measurable?

I have asked this question before on MSE, but have received no answer.

Answer by Michael Greinecker for From universal measurability to measurability

2013-05-23T01:39:36Z

I found a solution that suffices for what I do. It is based on strengthening the assumption that the graph $A$ is universally measurable to it being analytic. The notion of analyticity being used is that a subset $S$ of a measurable space $(M,\mathcal{M})$ is analytic if there is a compact metric space $T$ with Borel $\sigma$-algebra $\mathcal{B}(T)$ and a product measurable set $X\in\mathcal{M}\otimes\mathcal{B}(T)$ such that $S=\pi_M(X)$, where $\pi_M$ is the projection onto $M$. Analytic sets are always universally measurable. This notion of analyticity is extensively developed in the book Probability and Potential by Dellacherie and Meyer. A nice guide to the essentials can be found in this paper (JSTOR required).

The countable intersection or union of analytic sets is again analytic. If an analytic set is analytic in the product of some measurable space and a compact metric space, then the projection on the first coordinate is again analytic. An important fact from the theory of set-valued functions on a measurable space is that if the values are closed subsets of separable metric space, then the following condition is sufficient for the graph to be product measurable: The set $$\{\omega:C_\omega\cap O\neq\emptyset\}$$ is measurable for each open set $O$.

So assume that $A$ is analytic and $O$ is open. Then $$\{\omega:C_\omega\cap O\neq\emptyset\}=\pi_\Omega \big(A\cap(\Omega\times O)\big)$$ and hence analytic by the facts above and therefore universally measurable.

Answer by Michael Greinecker for Obtaining conditional probabilities as pushforwards of [0,1]

2013-04-18T05:29:08Z

I'am not sure this is what you want, but there is a way to represent regular conditional probabilities as some kind of pushforward-measure. Let $\kappa$ be a probability kernel from $Y$ to $X$. That is $\kappa:Y\times\mathcal{B}(X)\to[0,1]$ is measurable as a function of $Y$ and a probability measure as a function of $\mathcal{B}(X)$. Then there is a measurable function $f:Y\times[0,1]\to X$ such that $\kappa(y,\cdot)$ is the distribution of $f(y,\cdot)$. This can be found as Proposition 10.7.6 in Bogachev's Measure Theory.

Can random elements be defined in terms of a measure algebra?

2013-03-10T15:07:52Z

Let $(\Omega,\Sigma,\mu)$ be a probability space, $(X,\mathcal{X})$ be a measurable space and $R(\Omega,X)$ be the set of equivalence classes of measurable functions from $\Omega$ to $X$ under almost everywhere equality, they are random elements. Let $(\mathcal{A},\mu_A)$ be the measure algebra of $(\Omega,\Sigma,\mu)$, that is $\mathcal{A}$ identifies elements of $\Sigma$ if their symmetric difference has outer measure zero and $\mu_A$ is defined in the natural way in terms of its representatives.

I would like to know if one can identify the elements of $R(\Omega,X)$ with something that can be canonically be constructed in terms of $(\mathcal{A},\mu_A)$ and $(X,\mathcal{X})$.

The motivation behind the question is the following: I work with certain random elements that are defined on a countably generated probability space. By Maharam's theorem, this amounts to the measure algebra being isomorphic to one that consists of a convex combination of Lebesgue measure on $[0,1]$ and a discrete probability space. I would like to know whether it makes sense for me to say that I'm essentially working with such a probability space.

Can one view the Independent Product in Probability categorially?

2013-02-05T21:41:15Z

One can construct a category of probability spaces, but this category has no products. Now probability theory relies strongly on the ability to build independent products, the product measure. In a sense, the notion of independence is what distinguishes probablity theory from the theory of finite measures.

Is there a categorial way to make sense of and enlighten the notion of independent products in category theory?

It is possible to formulate independence in Lawvere's category of probabilistic mappings (Borel spaces as objects and Markov kernels as morphisms) in terms of constant morphisms, but I think this is not very enlightening, conditional independence is built into the morphisms. Maybe, this is what one has to do when putting probability center stage?

I do know the rudiments of categry theory, but I would prefer an answer that does not require too much immersion in category thory, provided that is possible.

Answer by Michael Greinecker for Is there a natural measures on the space of measurable functions?

2010-06-14T12:39:48Z

Let I be the unit interval with the Borel $\sigma$-algebra. There is no $\sigma$-algebra on the set of measurable functions from I to I such that the evaluation functional $e:I^I\times I\to I$ given by $e(f,x)=f(x)$ is measurable, as shown by Robert Aumann here, so even finding useful $\sigma$-algebras is a problem.

However, t is possible to talk about "almost all" functions in a function space even when it is not possible to have an appropriate measure. The trick is to find a characterization of a set having full (or zero) measure that can be applied to function spaces. There is a generalization of Lebesgue measure zero, independently found by various authors and knowyn as Haar measure zero or shyness that should be applicable to your problem. A nice survey of the theory and some of its extensions can be found here.

Answer by Michael Greinecker for Concise model of modern fiat money and its non-conservation

2013-01-27T10:44:37Z

I think an answer that discusses the actual institutional details of how the Fed controls the money supply would be off-topic here. Also, the Fed works slighlty differently from the ECB in that regard and there is more than one method of influencing the money supply (take a look at the wikipedia page on money creation). So I will try in this answer to demystify how a central bank can create money without literally sending out helicopters that drop fiat money on people.

First, one has to get right what money is. In explicit formal models, money is an asset that never pays out. If it has value, it is because there is a bubble in this asset. The first such model of money can probably be found in the 1958 paper An Exact Consumption-Loan Model of Interest with or without the Social Contrivance of Money by Paul Samuelson. It is worth pointing out that bubbles are not inherently bad and that paper constructs a toy economy in which everyone profits from the money bubble.

Now how can one increase the supply of an asset that never has to pay out anyways? It sells the asset in exchange for other assets. Since money never has to pay out, the central bank will not face a solvency constraint in the process. Selling money is not that different from selling milk, but since there are no cows involved, central banks are not constrained by cost.

Answer by Michael Greinecker for Topological conditions of Kolmogorov Extension Theorem

2013-01-11T16:04:37Z

The KET fails for general measurable spaces, the classical example can be found in a paper by Andersen and Jessen. Topological assumptions are necessary so that the resulting measure is not only finitely additive but countably additive. There exists a quasi-topological condition of measure spaces, perfectness, that is sufficient. A probability space $(\Omega,\sigma,\mu)$ is perfect if for every random variable $f:\Omega\to\mathbb{R}$, there exists a Borel set $B\subseteq f(\Omega)$ with measure one under the distribution $\mu\circ f^{-1}$. A proof of KET under the assumption that the marginal measures are perfect due to Lamb is given here. The strategy of the proof is to employ an existence result for regular conditional probability spaces and the construct the proces for them using the Ionescu-Tulcea theorem.

Answer by Michael Greinecker for Applications of the Giry monad in probability and statistics

2012-12-27T11:14:14Z

There is paper on the Arxiv, A categorical foundation for Bayesian probability by Culbertson and Sturtz. The paper contains also a very nice discussion of related literature.

Answer by Michael Greinecker for I know that you know...

2012-11-23T15:51:07Z

The classic model of these things is due to Robert Aumann and was introduced in his 1976 Agreeing to Disagree. There is a famous example due to Ariel Rubinstein, the electronic mail game, in which the behavior of $I_\omega$ radically differs from $I_n$ for any $n$.

Here is a way to show that one might have to apply this reasoning up to an arbitrarily large ordinal. Let $\alpha$ be a successor ordinal. Ann and Bob play the game of picking ordinals in $\alpha$ simultaneously. The one who chooses the highest number wins. One can never win by choosing $0$, so rationality rules out choosing $0$. It is however possible to win by choosing $1$ if the other player plays $0$. But if both Ann and Bob know that they are both rational, they have to choose at least $2$. It is clear that one has to iterate this reasoning up to the predecessor of $\alpha$.

Answer by Michael Greinecker for Mysterious sentence in a paper: what's the ultimate distribution of pure strategies?

2012-10-18T10:19:39Z

My interpretation: You identify a finite set of strategies $S$ with the unit vectors in $\mathbb{R}^{|S|}$, so that the simplex spanned becomes the space of mixed strategies. So every set of mixed strategies is a subset of this simplex. A probability distribution over $T$ induces again a probability distribution over $S$, so the set of probability distributions over the elements can again be identified with a subset of the simplex.

Answer by Michael Greinecker for Inequivalent complete norms and the axiom of choice

2012-10-01T15:54:02Z

No, it is not possible. There is a model due to Shelah of ZF+ dependent choice+every set of reals has the Baire property. There is a result of Garnir and Wright that implies that in such model, any two complete norms are equivalent. The Handbook of Analysis and Its Foundations by Eric Schechter has a chapter on this result and its consequences. The chapter is "The Dream Universe of Garnir and Wright". The result in that chapter is supposed to be slightly stronger than the one in the original papers, with which I'm not familiar.

Answer by Michael Greinecker for Nonstandard analysis in probability theory

2012-09-24T10:21:08Z

Non-standard analysis has been quite successful in settling existence questions in probability theory. Hyperfinite Loeb spaces allow for several constructions that cannot be done on standard probability spaces. In particular, NSA was quite useful for the construction of certain adapted processes. There is a paper by Hoover and Keisler, Adapted Probability Distributions, from 1984, in which the authors show that many of the properties that make hyperfinite Loeb spaces so useful where due to a property they called saturation: A probability space $(\Omega,\Sigma,\mu)$ is saturated if whenever $\nu$ is a Borel probability measure on $[0,1]^2$ and $f:\omega\to[0,1]$ a random variable with distribution equal to the marginal of $\nu$ on the first coordinate, then there exists a random variable $g:\Omega\to[0,1]$ such that the distribution of $(f,g)$ is $\nu$. An example of a saturated probability space that is not a hyperfinite Loeb space is the coin-flipping measure on $\{0,1\}^\kappa$ when $\kappa$ is uncountable. A relatively readable exposition of this approach can be found in the small book Model Theory of Stochastic Processes by Fajardo and Keisler. There are also several related papers and surveys on Keisler's homepage.

In a sense, we nowadays understand fairly well how certain powerful techniques of non-standard analysis work below the surface, so we can use a lot of the constructions freed of NSA. There isn't really anything where it is necessary to use NSA. Still, NSA is a rather powerful and useful tool. A good overview over what it can do for probability theory, mainly the theory of sochastic processes, is in the article by Osswald and Sun in Nonstandard Analysis for the Working Mathematician by Loeb.

Answer by Michael Greinecker for Reference request: an elementary proof of Brouwer fixed-point theorem.

2012-09-24T13:44:39Z

Here are two further references of proofs of the fixed point theorem that rely on evaluating determinants:

MR0117523 Dunford, Nelson ; Schwartz, Jacob T. Linear Operators. I. General Theory. With the assistance of W. G. Bade and R. G. Bartle. Pure and Applied Mathematics, Vol. 7 Interscience Publishers, Inc., New York; Interscience Publishers, Ltd., London 1958 xiv+858 pp. The proof is on page 467.

MR0610487 Kannai, Yakar . An elementary proof of the no-retraction theorem. Amer. Math. Monthly 88 (1981), no. 4, 264--268.

Answer by Michael Greinecker for Symmetries of probability distributions

2012-09-05T12:52:12Z

Usually, such symmetries have been either studied in the context of Lebesgue spaces or studied in the context of homogenous measure algebras, where autmorphisms are easy to study. Every automorphism of a probability space gives rise to an automorphism of the corresponding measure algebra.

The easiest case are Lebesgue spaces, or even simpler, studying the uniform distribution on $[0,1]$. This is of course essentially the case of a Haar measure. A nice property is that one can take any automorphism of the measure algebra and find an automorphism of the probability space inducing it. Moreover, two automorphisms of the probability space giving rise to the same automorphism of the measure algebra can differ only on a set of measure zero. Every, homogenous, atomless, separable measure algebra can be represented as a Lebesgue space.

If one starts with a homogenous measure algebra, one may look for probability spaces representing the measure algebra. The two most prominent representations are by the Stone space of the measure algebra or in the form of a product of coinflips $\{0,1\}^\kappa$ with $\kappa$ infinite, which can represent every atomless (normed) homogenous measure algebra by Maharam's theorem. In the case of the Stone space, the automorphisms of the measure algebra correspond essentially to the automorphisms of the representing probability space. In the coin-flipping representation, every automorphism of the measure algebra is induced by an automorphism of the probability space. But very different automorphisms may give rise to the same automorphism of the measure algebra. Actually, there exists an automorphism of $\{0,1\}^\mathfrak{c}$ that induces the identity on the measure algebra but has no fixed point.

The discussion so far is largely adapted from the introduction to Ergodic theory on homogeneous measure algebras. by Choksi and Prasad. The book this has appeared in is likely to be available somewhere on the internet...

One can also study the case of rigid probability spaces, where there is essentially no automorphism but the identity. It is actually possible to find a countably generated and countably separated measurable space in which all automorphisms differ from the identity only on a countable set. This is done in the booklet Borel Spaces by Rao and Rao in Proposition 4. There also is an example of an atomless, countably generated probability space with no autmorphism but the identity (up to a countable set) in Section 48 of Values of non-atomic games by Aumann and Shapley.

Answer by Michael Greinecker for Simple functions in the product space

2012-09-03T13:26:51Z

All indicator functions of the algebra generated by rectangles are linear combinations of indicator functions of rectangles. Indicator functions of product measurable sets can be approximated by indicator functions of the algebra, which follows from the standard Caratheodory construction. Clearly, the same can be done with linear combinations of indicator functions.

Answer by Michael Greinecker for Nash Equilibrium of simple betting game

2012-08-20T14:10:33Z

Outside Lake Wobegon, not all players can receive above average values. Intuitively, what a player does when betting can only matter when the other player bets too. But if the player knows that the other player is betting, the player knows that the other player is confidet not to lose- which should dampen your confidence. UNless you are really confidemt, but then the other player knows you only bet because you are really confident, which should dampen her confidence...

The phenomenon is well known an related to the literature on the impossibility of common knowledge agreeing to disagree under a common prior, as pioneered by Robert Aumann.

You can simplify the analysis by assuming that there is a natural number $n$ and each players number is uniformly and independently drawn from $\{1,2,\ldots,n\}$. For simplicity, assume that both players get $0$ under a draw. Take any Nash equilibrium. Let $m$ be the smallest number such that any player who has drawn $m$ would bet. Suppose $m< n$. If there is a player willing to bet at $m$, it is the unique best response of the other player to play when getting $n$. But this means that the player betting with $m$ can never win but potentially lose with positive probability, which gives a worse value than folding. Since everyone is supposed to play a best response to the equilibrium strategy profile, we get a contradiction, players never bet with values smaller than $n$.

Answer by Michael Greinecker for The (Sigma) Algebra of Convex Sets

2012-08-17T09:10:57Z

The answer to the second question is yes. Trivially, every closed convex set is closed and hence in the Borel $\sigma$-algebra as the complement of an open set.

For the other direction, note that every open or closed ball is convex. Every open ball is an incrasing union of closed balls. And every open set is a countable union of open balls, since $\mathbb{R}^n$ is separable.

If one wants to restrict the problem to a compact, convex subset of $\mathbb{R}^n$, one can use the fact that a trace $\sigma$-algebra is generated by the traces of generators. That is, if $(Y,\mathcal{Y})$ is a measurable space and $\mathcal{G}\subseteq 2^Y$ such that $\mathcal{Y}=\sigma(\mathcal{G})$, and $X\subseteq Y$, then the $\sigma$-algebra $\{B\cap X:B\in\mathcal{Y}\}$ equals $\sigma\big(\{X\cap G:G\in\mathcal{G}\}\big)$.

Answer by Michael Greinecker for "Nice" sigma-algebra on set of measurable functions

2012-08-13T23:47:36Z

There is an impossibility theorem: If you let $\mathcal{L}$ be the the space of Borel-measurable functions $f:[0,1]\to[0,1]$, and $e:\mathcal{L}\times [0,1]\to[0,1]$ the evaluation given by $e(f,x)\mapsto f(x)$, then there is no $\sigma$-algebra on $\mathcal{L}$ such that the evaluation is jointly measurable. The result is a consequence of the rather complicated classification result in R. Aumann, Borel Structures for Function Spaces, Illinois Journal of Mathematics 5 (1961), pp. 614-630. Easier proofs of the main results can be found in the paper "Borel Structures for Function Spaces" (yes, same title) by B.V. Rao, Colloquium Mathematicum, 1971.

A $\sigma$-algebra on measurable functions I have actually seen used is the following: If $(S,\mathcal{S})$ and $(T,\mathcal{T})$ are measurable spaces, we endow the family of measurable functions between them with the $\sigma$-algebra generated by sets of the form $\{f:f(s)\in B\}$ with $s\in S$ and $B\in\mathcal{T}$. The author used this $\sigma$-algebra to show that to each Markov kernel from $S$ to $T$, there corresponds a certain probability measure on this $\sigma$-algebra. The paper is H. v. Weizsäcker Zur Gleichwertigkeit zweier Arten der Randomisierung, Manuscripta Mathematica 11 (1974).

Random Functions and Transition Probabilities

2011-12-24T17:46:57Z

Let $(S,\mathcal{S})$ and $(T,\mathcal{T})$ be measurable spaces. A transition probability from $S$ to $T$ is a function $\pi:S\times\mathcal{T}\to [0,1]$ such that $\pi(s,\cdot)$ is a probability measure for all $s\in S$ and $\pi(\cdot,B)$ is measurable for all $B\in\mathcal{T}$.

Now let $(\Omega,\Sigma,\mu)$ be a probability space and let $f:\Omega\times S\to T$ be a jointly measurable function.

Under what conditions is $\pi_f:S\times\mathcal{T}\to [0,1]$ given by $\pi_f(s,B)=\mu \{\omega:f(s,\omega)\in B\}$ a transition probability?

Clearly, the issue is measurability.

Motivation: If $(T,\mathcal{T})$ is standard Borel, there always exists a function such that $\pi=\pi_f$ when $\Omega$ is atomless, this is Proposition 10.7.6. in Bogachev. Moreover, such random functions have been used as a definition of mixed strategies in game theory and I think it would be interesting to understand them in terms of transition probabilities in a general setting.

Answer by Michael Greinecker for Random Functions and Transition Probabilities

2012-07-26T13:17:56Z

I've found a rather simple solution. No additional assumptions are necessary. There is a natural method of composing transition probabilities that gives rise to a new transition probability (see for example (2) here).

We can identify probability measures with transition probabilities constant in the first argument. Also, we can identify $f$ with the transition probability $\kappa:\Omega\times S\times\mathcal{T}\to[0,1]$ given by $\kappa(\omega,s,B)=1$ if $f(\omega,s)\in B$ and $\kappa(\omega,s,B)=0$ if $f(\omega,s)\notin B$.

Also, the pointwise product of transition probabilities is a transition probability. So let $\kappa_\mu:S\times\Sigma\to[0,1]$ the representation of $\mu$ as a tranition probability. Also, let $\kappa_1:S\times\mathcal{S}\to[0,1]$ be the identity transition probability on $S$. Let $\pi:S\times\mathcal{S}\otimes\mathcal{T}\to[0,1]$ be the pointwise product of $\kappa_\mu$ and $\kappa_1$.

Then $\pi_f=\kappa\circ\pi$ and is therefore a transition probability as the composition of transition probabilities.

Answer by Michael Greinecker for Is a random subset of the real numbers non-measurable? Is the set of measurable sets measurable?

2012-07-17T00:12:59Z

The problem of choosing subsets at random has been studied in a rather different context in mathematical economics. Suppose we choose a subset of $[0,1]$ by independently throwing a fair coin for each number. Heuristically, such a set should have measure $1/2$. For what we do is randomly choose an indicator function with pointwise expectation $1/2$. By some intuitive apeal to a law of large numbers, the sample realizations should have the same expectation. This kind of reasoning is widely used in economics. A large population is modeled by a continuum and even when each person faces individual uncertainty, there should be no aggregate uncertainty.

For the reason given by Will Sawin, the naive approach doesn't work quite well. For Lebesgue measure, some intuition comes from Lusin's theorem to the effect that every measurable function is continuous on a "large" subset. Continouity is a condition to the effect that the value at a point is closely related to the value at nearby points. If you choose independently at each value, you wouldn't expect to get a function continuous on a large set.

The general tradeoff between independence and measurable sample realizations is strongly expressed in the following result of Yeneng Sun:

Proposition: Let $(I,\mathcal{I},\mu)$ and $(X,\mathcal{X},\nu)$ be probability spaces with (complete) product probability space $(I\times X,\mathcal{I}\otimes\mathcal{X},\mu\otimes\nu)$ and $f$ be a jointly measurable function from $I\times X$ to $\mathbb{R}$ such that for $\mu\otimes\mu$-almost all $(i,j)$ the functions $f(i,\cdot)$ and $f(j,\cdot)$ are independent. Then for $\mu$-almost all $i$, the function $f(i,\cdot)$ is constant.

Note that the independence condition in this result is quite weak. Sun calls it almost sure pairwise independence. But an important discovery by Sun was that if joint measurability and almost sure pairwise independence were compatible, one could obtain an exact law of large numbers for a continuum of random variables by an application of Fubini's theorem. In particular, such a law of large numbers holds for extensions of the product spaces that allow for the conclusion of Fubini's theorem to hold and still allow for nontrivial (a.s. pairwise) independent processes. He called such extensions rich Fubini extensions and gave one example of such a product space: The Loeb product of two hyperfinite Loeb spaces. So one can get natural random sets for some spaces. The reference is: The exact law of large numbers via Fubini extension and characterization of insurable risks (2006)

A systematic study of rich Fubini extensions was done by Konrad Podczeck in the paper On existence of rich Fubini extensions (2010), in which he has essentially shown that one can choose random subsets of a probability space if and only if the probability space has the following property, which he called super-atomlessnes (and which is known by a lot of other names such as saturation):

For any subset $A$ with positive measure, the measure algebra of the trace on $A$ does not coincide with the measure algebra of a countably generated space.

Lebesgue measure on the unit interval does not satisfy this condition, but there exists extensions of Lebesgue measure that are superatomless.

Conclusion: One cannot obtain random Lebesgue measurable sets in a sensible way by choosing independently elements, but one can choose random sets in an extension of Lebesgue measure this way.

Answer by Michael Greinecker for Applied Problems in Probability which can not be modelled on Polish spaces

2012-07-10T19:14:26Z

This answer is identical to the one I gave when the same question was posted at M.SE.:

There are a number of constructions that do not work for Polish spaces, but a certain class of probability spaces, variously known as super-atomless, saturated, nowhere countably generated and a number of other names. A nice overview can be found here.

A probability space $(\Omega,\Sigma,\mu)$ is saturated if for every two Poilsh spaces $X$ and $Y$, every probability measure $\nu$ on $X\times Y$ and every random variable $f:\Omega\to X$ such that its distribution $\mu f^{-1}$ equals the marginal of $\nu$ on $X$, there is a random variable $g:\Omega\to Y$ such that the joint distribution of $(f,g)$ is $\nu$.

The following definition is conceptually different, but can be shown to be equivalent:

A probability space $(\Omega,\Sigma,\mu)$ is super-atomless if there is no $A\in\Sigma$ satisfying $\mu(A)>0$, such that the pseudo-metric space obtained by endowing the trace $\sigma$-algebra on $A$ with the pseudo-metric $d(A,B)=\mu(A\triangle B)$ is separable.

Answer by Michael Greinecker for Blackbox Theorems

2012-06-14T07:02:27Z

The Borel isomorphism theorem says that any two Polish (complete and separable metrizable) spaces endowed with their Borel $\sigma$-algebra are isomorphic as measurable spaces if and only if they have the same cardiality and this cardinality is either countable or the cardinality of the continuum.

The result is extremely useful and widely applied in probability theory. It allows one to prove many results for general Polish spaces by proving them for the real line or the unit interval. The proof is actually not that hard, but somewhat messy and gives little useable insight for those not working in descriptive set theory.

Atoms of a sequence of Sigma-algebras

2012-05-24T13:44:55Z

I'm trying for some time now to prove or disprove the following conjecture to no avail:

Let $S$ be a set and let $(\Sigma _n)$ be a sequence of countably generated $\sigma$-algebras on $S$ satisfying the following two conditions:

$\Sigma_n\subseteq\Sigma_{n+1}$ for all $n$.

If $A\in\Sigma_{n+1}$ is a union of $\Sigma_n$-atoms, then $A\in\Sigma_n$ for all $n$.

Then for all $n$: If $A\in\sigma\big(\bigcup_n\Sigma_n\big)$ is a union of $\Sigma_n$-atoms, then $A\in\Sigma_n$.

An atom is a minimal measurable set. In a countably generated $\sigma$-algebra, the atoms form a partition of the underlying space into points that can not be distinguished by measurable sets.

I have actually only little intuition for the problem. If $S$ is analytic and all the $\Sigma_n$ are sub-$\sigma$-algebras of the Borel-$\sigma$-algebra, both condition 2. and the conjecture is automatically satisfied, due to a result of Blackwell, so counterexamples must be somewhat unnatural.

Answer by Michael Greinecker for Is there a good comparative study of the Banach integral?

2012-05-21T12:45:43Z

One can find some information in the German language book Reelle Zahlen by Oliver Deiser. Banach apparently introduced his integral in the paper Sur le problème de la mesure, Fund. Math. 4, 1923. It was apparently introduced to show that a translation invariant and finitely additive extension of Lebesgue measure on all sets of real numbers exist.

The Banach integral of a Riemann-integrable function coincides with the Riemann integral. There is a function whose Banach integral is $0$ that is not Lebesgue integrable. Also there are Lebesgue integrable functions whose Banach integral differs from The Lebesgue integral. The Banach integral is linear, but not "countable additive". The last fact explains that one cannot do all the nice limiting arguments one is used to when working with the Lebesgue integral.

Answer by Michael Greinecker for Mathematical analysis of Lewisian concepts, esp. natural properties

2012-04-23T21:47:32Z

Lewis introduced the concept of common knowledge (I know that you know that I know that you know that...) in his book Convention. The concept has been formalized in a partitional model of knowledge in the simple and elegant paper Agreeing to Disagree by Robert Aumann. Aumann wasn't aware of the prior work of Lewis. The concept of common knowledge became one of the building blocks of modern non-cooperative game theory and has been extensively studied and generalized. A survey can be found here.

Answer by Michael Greinecker for Maximal ideals in Boolean algebras; reference request

2012-04-05T12:29:10Z

The result on the number of maximal ideals can be found as Corollary 7.4. in the book Comfort & Negrepontis "The Theory of Ultrafilters", 1974. A proof can also be found at PlanetMath here.

Unique Existence and the Axiom of Choice

2011-11-16T15:51:01Z

The axiom of choice states that arbitrary products of nonempty sets are nonempty. Clearly, we only need the axiom of choice to show the non-emptiness of the product if there are infinitely many choice functions. If we use a choice function to construct a mathematical object, the object will often depend on the specific choice function being used. So constructions that require the axiom of choice often do not provide the existence of a unique object with certain properties. In some cases they do, however. The existence of a cardinal number for every set (ordinal that can be mapped bijectively onto the set) is such an example.

What are natural examples outside of set theory where the existence of a unique mathematical object with certain properties can only be proven with the axiom of choice and where the uniqueness itself can be proven in ZFC (I don't want the uniqueness to depend on a specific model of ZFC)?

The next question is a bit more vague, but I would be interested in some kind of birds-eye view on the issue.

Are there some general guidelines to understand in which cases the axiom of choice can be used to construct a provably unique object with certain properties?

This question is motivated by a discussion of uniqueness-properties of certain measure theoretic constructions in mathematical economics that make heavy use of non-standard analysis.

Edit: Examples so far can be classified in three categories:

Cardinal Invariants: One uses the axiom of choice to construct a representation by some ordinal. Since ordinals are canonically well ordered, this gives us a unique, definable object with the wanted properties. Example: One takes the dimension (as a cardianl) of a vector space and constructs the vector space as functions on finite subsets of the cardinal (François G. Dorais).

AC Properties: One constructs the object canonically "by hand" and then uses the axiom of choice to show that it has a certain property. Trivial example: $2^\mathbb{R}$ as the family of well-orderable sets of reals.

Employing all choice functions: Here one gets uniqueness by requiring the object to contain in some sense all objects of a certain kind that can be obtained by AC. Examples: The Stone-Čech compactification as the set of all ultrafilters on it (Juris Steprans), or the dual space of a vector space, the space of all linear functionals. (Martin Brandenburg) The AC is used to show that these spaces are rich enough. Formally, these examples might be categorized in the second category, but they seem to have a different flavor.

Answer by Michael Greinecker for Reference for the Brouwer fixed point theorem

2012-03-06T10:53:04Z

The general theorem was first given in:

Brouwer, L. E. J. Über Abbildung von Mannigfaltigkeiten. Math. Ann. 71, 97-115. Berichtigung ebd. S. 598 (1912).

Comment by Michael Greinecker

2013-05-23T22:43:14Z

There are much more qualified people to comment here, but if I recall correctly, it was extensively discussed whether to include categories or not.

Comment by Michael Greinecker

2013-05-13T11:40:48Z

Yes, with "finite" replaced by "measurable". Dellarcherie and Meyer do this in this generality.

Comment by Michael Greinecker

2013-04-21T12:07:40Z

They are not equivalent.

Comment by Michael Greinecker

2013-04-21T11:31:43Z

Not all definitions of regular measures are equivalent, so you should be more explicit with what the "standard definition" is. Also, this is not really research mathematics, so you might want to ask this where such questions are appropriat, such as math.stackexchange.com

Comment by Michael Greinecker

2013-04-12T19:13:16Z

I think you might look cooler if you can relate to topics outside math.

Comment by Michael Greinecker

2013-04-10T09:33:09Z

This should certainly be community wiki. This thread is closely related and gives examples of shorter papers: mathoverflow.net/questions/7330/…

Comment by Michael Greinecker

2013-03-16T19:10:56Z

I don't think you've came to the right place. math.stackexchange.com will be of more use t you in getting an answer to this question.

Comment by Michael Greinecker

2013-03-05T12:51:49Z

I once asked a related question: mathoverflow.net/questions/81082/…

Comment by Michael Greinecker

2013-03-01T01:42:14Z

The problem is essentially equivalent to the Brouwer fixed point theorem (at least without gross substitutes).

Comment by Michael Greinecker

2013-02-24T14:34:17Z

Most proofs of the Daniell-Kolmogorov consistency theorem consist of two steps: Showing that a finitely additve measure on the field is induced. Using a regularity argument to show it is sigma-additive. So, there should be lots of proofs you could cite.

Comment by Michael Greinecker

2013-02-24T02:09:09Z

And what should a product measure satisfy? Fubini's theorem does not hold for finitely additive probabilities.

Comment by Michael Greinecker

2013-02-24T00:23:06Z

On what sets should the product probability be defined?

Comment by Michael Greinecker

2013-02-19T18:27:25Z

@Rabee He explains how to prove the general result from existence for finite games using narrow convergence. That's not really practical for classroom use, but should convince mathematicians.

Comment by Michael Greinecker

2013-02-13T11:23:58Z

This question is not really on topic here. Mathoverflow specializes in research level questions. This focus is both its strength and its weakness, depending on your interests and needs. A better outlet for such a question would be math.stackexchange.com.

Comment by Michael Greinecker

2013-02-12T11:11:52Z

The link seems to be broken now.